Looking for classes? Ksquare Career Institute, Bengaluru →
The charge following through a resistance R varies with time t as $Q = at - bt^2$, where a and b are positive constants. The total heat produced in R is
A
$\frac{a^3R}{3b}$
B
$\frac{a^3R}{2b}$
C
$\frac{a^3R}{b}$
D
$\frac{a^3R}{6b}$
Explanation
Integrate $I^2R$ from t = 0 until the current becomes zero.
Detailed Solution
Current $I = \frac{dQ}{dt} = a - 2bt$; I = 0 at $t = \frac{a}{2b}$.
$dW = I^2R\,dt$
$W = \int_0^{a/2b}(a - 2bt)^2R\,dt = R\left[a^2t + \frac{4b^2t^3}{3} - \frac{4abt^2}{2}\right]_0^{a/2b}$
$W = R\left[\frac{a^3}{2b} + \frac{a^3}{6b} - \frac{a^3}{2b}\right] = \frac{Ra^3}{2b}\times\frac{1}{3}$
$W = \frac{Ra^3}{6b}$
$dW = I^2R\,dt$
$W = \int_0^{a/2b}(a - 2bt)^2R\,dt = R\left[a^2t + \frac{4b^2t^3}{3} - \frac{4abt^2}{2}\right]_0^{a/2b}$
$W = R\left[\frac{a^3}{2b} + \frac{a^3}{6b} - \frac{a^3}{2b}\right] = \frac{Ra^3}{2b}\times\frac{1}{3}$
$W = \frac{Ra^3}{6b}$